Quantitative Aptitude Test

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Quantitative Aptitude Test - 13

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Weekly Quiz Competition

Question 1
1 / -0.25

When the arrangements of a two-digit number are interchanged, then the number that occurs is smaller than the original number by 27. The ratio of the digits of the number is 2 : 1. Determine the original number?

A
24

B
36

C
48

D
63

Solution

Assume that the two-digit number is 10x + y.
So, according to the question,
10x + y – (10y + x) = 27
or 9x – 9y = 27
or x – y = 3
Again,
x = 2k and y = k
Thus,
x = 2 $\times$ 3
= 6
Therefore, the required number is (10 $\times$ 6 + 3) = 63.
Hence, the correct option is (d).
Question 2
1 / -0.25

A mixture of wheat is sold at Rs. 6 per kg. This mixture contains wheat qualities of prices Rs. 4.5 per kg and Rs. 5.3 per kg. For a profit of 20%, what should be the ratio of cheaper quality to costlier quality of wheat?

A
5 : 3

B
5 : 2

C
1 : 5

D
3 : 5

Solution

Let the cost price of the mixture be Rs. ' $a^{\prime}$ per kg.
Selling price of the mixture $=$ Rs. $(a+0.20 a)$
$\Rightarrow 120 \mathrm{a}=600$
$\Rightarrow a=5$
Now, average cost price of the mixture of wheat $=R s .5$ per $k g$
Suppose $1 \mathrm{kg}$ of mixture contains 'y' kg of cheaper wheat quality and $(1-\mathrm{y}) \mathrm{kg}$ of costlier wheat quality. Then, $4.5 y+5.3(1-y)=5$
$4.5 y+5.3-5.3 y=5$
$0.3=0.8 y$
$\Rightarrow \mathrm{y}=\frac{3}{8}$
and, $1-y=\frac{5}{8}$
Hence, ratio of cheaper quality to costlier quality of wheat $=\frac{3}{8}: \frac{5}{8}=3: 5$
Hence, the correct option is (d).
Question 3
1 / -0.25

In a group of junior basketball players, the average weight of boys is 65.9 kg and the average weight of girls is 57 kg. When the average weight of both boys and girls is 60.3 kg and the total number of boys in the group is 66, then what is the total number of girls in the group?

A
112

B
154

C
162

D
168

Solution

Average weight of boys is $65.9 \mathrm{kg}$.
Average weight of girls is 57 kg.
Average weight of all players is $60.3 \mathrm{kg}$.
Let us assume that the total number of players is 'x'. Then,
$\frac{65.9 \times 66+(x-66) \times 57}{x}=60.3$
$65.9 \times 66+57 x-57 \times 66=60.3 x$
or $(65.9-57) \times 66=3.3 \mathrm{x}$
or $8.9 \times 66=3.3 \mathrm{x}$
$178=x$
Thus, the number of girls is $178-66=112$.
Hence, the correct option is (a).
Question 4
1 / -0.25

Directions For Questions

The bar graph given below shows the sales of cars (in thousand numbers) from six automobile companies (M1, M2, M3, M4, M5 and M6) during the two successive years 2014 and 2015. Answer the question (in thousand numbers) based on the given information.

...view full instructions

What are the total sales achieved by the companies M1, M3 and M5 together, in the years 2014 and 2015?

A
250,000

B
310,000

C
435,000

D
560,000

Solution

Total sales $=\mathrm{M}_{1}+\mathrm{M}_{3}+\mathrm{M}_{5}$

$=(80+105)+(95+110)+(75+95)$

$=560$

Hence, the correct option is (d).
Question 5
1 / -0.25

Directions For Questions

The bar graph given below shows the sales of cars (in thousand numbers) from six automobile companies (M1, M2, M3, M4, M5 and M6) during the two successive years 2014 and 2015. Answer the question (in thousand numbers) based on the given information.

...view full instructions

M₆’s total sale for the years 2014 and 2015 is what percent of M₃’s total sales for the same years?

A
73.17%

B
71.11%

C
69.53%

D
72.51%

Solution

M₆’s total sale for the years 2014 and 2015 is 70 and 80 respectively.

M₃’s total sale for the years 2014 and 2015 is 95 and 110 respectively.

Required percentage $=\left[\frac{(70+80)}{(95+110)} \times 100\right] \%$

$=\left(\frac{150}{205} \times 100\right) \%$

$=73.17 \%$

Hence, the correct option is (a).
Question 6
1 / -0.25

Directions For Questions

The bar graph given below shows the sales of cars (in thousand numbers) from six automobile companies (M1, M2, M3, M4, M5 and M6) during the two successive years 2014 and 2015. Answer the question (in thousand numbers) based on the given information.

...view full instructions

What is the ratio of M2’s total sales for the years 2014 and 2015 to M4’s total sales for the same years?

A
3 : 2

B
5 : 7

C
7 : 9

D
3 : 4

Solution

M₂’s total sales for the years 2014 and 2015 is 75 and 65 respectively.
M₄’s total sales for the years 2014 and 2015 is 85 and 95 respectively.
$\text {Required ratio } $
$=\frac{(75+65)}{(85+95)}$
$=\frac{140}{180}$
$=\frac{7}{9}$
$=7: 9 $
Hence, the correct option is (c).
Question 7
1 / -0.25

Directions For Questions

The bar graph given below shows the sales of cars (in thousand numbers) from six automobile companies (M1, M2, M3, M4, M5 and M6) during the two successive years 2014 and 2015. Answer the question (in thousand numbers) based on the given information.

...view full instructions

What are the average sales of all the six automobile companies (in thousand numbers) for the year 2014?

A
83

B
80

C
82

D
84

Solution

Total sales of M1, M2, M3, M4, M5, M6 automobile companies (in thousand numbers) for the year 2014 are 80, 75, 95, 85, 75, 70 respectively.
$\text { Average sales }
=\frac{1}{6} \times(80+75+95+85+75+70) $
$=\frac{1}{6} \times(480) $
$=80 $
Hence, the correct option is (b).
Question 8
1 / -0.25

Directions For Questions

The bar graph given below shows the sales of cars (in thousand numbers) from six automobile companies (M1, M2, M3, M4, M5 and M6) during the two successive years 2014 and 2015. Answer the question (in thousand numbers) based on the given information.

...view full instructions

The average sales of companies M1, M3 and M6 in 2014 are what percent of the average sales of companies M1, M2 and M3 in 2015?

A
87.5%

B
82.5%

C
83.6%

D
81.3%
Question 9
1 / -0.25

The interest on a certain principal P at 12% SI for 4 years is equal to the amount on a principal Q at 11% S.I. for 4 years. Then:

A
12P = 11Q

B
11P = 12Q

C
P = 3Q

D
P = 4Q

Solution

l₁ = (P)

A₂= Q + Q

Now, l₁ = A2

Or, 48P = 100Q + 44Q

P = 3Q

Hence, the correct option is (c).
Question 10
1 / -0.25

1 . 2 . 3 + 2 . 3 . 4 + 3 . 4 . 5 + ………………. + n (n + 1) (n + 2) is the same as :

A

$n^{2}(n+1)^{\frac{2}{4}}$

B

$\frac{n(n+1)(2 n+1)}{6}$

C

$\frac{n(n+1)(n+2)(n+3)}{4}$

D

$\mathrm{n}(\mathrm{n}+1)$

Solution

Substitute $n=1,2, \ldots$ and eliminate options.
Note: In case an option is 'none of these', check for three consecutive values of "n'.
Alternatively, $n^{\text {th }}$ term is $n(n+1)(n+2)=n^{3}+3 n^{2}+2 n$
Recall: $\sum n^{3}=[\frac{(n+1)}{2}]^{2} ; \sum n^{2}=\frac{n(n+1)(2 n+1)}{6} ; \sum n=\frac{n(n+1)(n+2)(n+3)}{4}$
$S_{n}=\sum\left(n^{3}+3 n^{2}+2 n\right)=\sum n^{2}+3 . \sum n^{2}+2$.
$\sum n=\frac{n(n+1)(n+2)(n+3)}{4}$
Hence, the correct option is (c).

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Re-Attempt Weekly Quiz Competition

Quantitative Aptitude Test - 13

Result

Quantitative Aptitude Test - 13

Score

-

Rank

-

SHARING IS CARING

Question 1

24

36

48

63

Solution

Question 2

5 : 3

5 : 2

1 : 5

3 : 5

Solution

Question 3

112

154

162

168

Solution

Question 4

250,000

310,000

435,000

560,000

Solution

Question 5

73.17%

71.11%

69.53%

72.51%

Solution

Question 6

3 : 2

5 : 7

7 : 9

3 : 4

Solution

Question 7

83

80

82

84

Solution

Question 8

87.5%

82.5%

83.6%

81.3%

Question 9

12P = 11Q

11P = 12Q

P = 3Q

P = 4Q

Solution

Question 10

\(n^{2}(n+1)^{\frac{2}{4}}\)

\(\frac{n(n+1)(2 n+1)}{6}\)

\(\frac{n(n+1)(n+2)(n+3)}{4}\)

\(\mathrm{n}(\mathrm{n}+1)\)

Solution

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